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    <script>
      // 利用递归函数求1~n的阶乘 1 * 2 * 3 * 4 * ..n
      function fn(n) {
        if (n == 1) {
          return 1
        }
        return n * fn(n - 1)
      }
      console.log(fn(3))
      console.log(fn(4))
      // 详细思路 假如用户输入的是3
      //return  3 * fn(2)
      //return  3 * (2 * fn(1))
      //return  3 * (2 * 1)
      //return  3 * (2)
      //return  6
    </script>
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